LEC.6 (1)

Intravenous
Infusion- part 1
Intravenous (IV) drug solutions may be either
injected as a bolus dose (all at once) or infused
slowly through a vein into the plasma at a constant
rate (zero order ) over extended time.
IV bolus injection IV infusion

The main advantage for giving a drug by IV
infusion:
 it allows precise control of plasma drug
concentrations to fit the individual needs of the
patient
 It maintains an effective constant
plasma drug concentration by
eliminating wide fluctuations
between the peak (maximum) and
trough (minimum) plasma drug
concentration.
 the duration of drug therapy may be maintained or

terminated as needed using IV infusion
One compartment open model – IV infusion
R0 - KE
Drug Blood and other Elimination
Zero- order body tissue
Infusion rate
The change in the amount of drug in the body at any time

(dD/dt) during the infusion is the rate of input (R0= infusion
rate) minus the rate of output (KE D= elimination rate)
dD/dt= Ro - KE D
, integration and rearrangement
D= (R0/KE) (1-e-k t) E D = Cp VD
ClT= K VD
Cp is the plasma drug concentration at any time during the IV

infusion, where t is the time for infusion
Steady-State drug concentration (Css) and time needed to reach Css
Figure: The plasma drug concentration-time curve of a drug given by constant IV infusion
At steady state, the rate of drug eliminaton from the body is

equal to the rate of drug (infusion rate) entering the body.
Therefore, at steady state, the rate of change in the plasma
drug concentration dCp/dt = 0
dCp/dt = R0 – KE (Css/VD) , zero= R0 – KE (Css/ VD )
Css= R0 / KE VD
Steady-State drug concentration (Css) and time needed to reach Css
 No net change in the amount of drug in the body, DB, as a

function of time during steady state
Whenever the infusion stops, either before

or after steady state is reached, the drug
concentration always declines
exponentially according to (first-order)
kinetics. The slope of the elimination
curve equals to -k/2.3
For a zero-order elimination processes, if rate of input is

greater than rate of elimination, plasma drug concentrations
will keep increasing and no steady state will be reached
 The plasma drug concentration at steady state (Css) is
related to the rate of infusion and inversely related to the
body clearance of the drug
 In clinical practice, the drug activity will be observed

when the drug concentration is close to the desired plasma
drug concentration, which is usually the target or desired
steady-state drug concentration.
 The time to reach 90%, 95%, and 99% of the steady-state
drug concentration, Css, may be calculated
Calculate the time need to reach 99% of the steady state
concentration
The time for a drug whose t1/2 is 6 hours to reach 95% of the
steady-state plasma drug concentration will be 5 t1/2, or 5 × 6
hours = 30 hours.
The time to reach steady-state

concentration (Css) is dependent
upon the elimination half-life (t1/2)
and not infusion rate
An increase in the infusion rate will not shorten the time

to reach the steady-state drug concentration. If the drug is
given at a more rapid infusion rate, a higher steady-state
drug level will be obtained, but the time to reach steady state
is the same.
Example - An antibiotic has a volume of distribution (VD) of
10 L and a k of 0.2 h–1. A steady-state plasma concentration
(Css) of 10 μg/mL is desired. The infusion rate needed to
maintain this concentration can be determined as follows:
R = Css VD k = (10 μg/mL) (10) (1000 mL) (0.2 h–1) = 20 mg/h
If the patient has a uremic condition and the elimination rate

constant has decreased to 0.1 h–1. To maintain the steady state
concentration of 10 μg/mL, need to determine a new rate of
infusion as follows.
R = (10 μg/mL) (10) (1000 mL) (0.1 h–1) = 10 mg/h
When the elimination rate constant decreases, then the infusion
rate must decrease proportionately to maintain the same Css.
However, because the elimination rate constant (k) is smaller
(i.e.; the elimination t1/2 is longer), the time to reach Css will be
longer.
Example - A patient was given an antibiotic (t1/2 = 6 hours) by
constant IV infusion at a rate of 2 mg/h. At the end of 2 days,
the serum drug concentration was 10 mg/L. Calculate the total
body clearance ClT for this antibiotic.
Css = R / Cl
Cl = R/ Css
The serum sample was taken after 2 days or 48 hours of
infusion, which time represents 8 × t1/2, therefore, this serum
drug concentration approximates the Css
Infusion method for calculating patient elimination half–life:
Example - An antibiotic has an elimination half-life of 3-6 hours
in the general population. A patient was given an IV infusion of
an antibiotic at an infusion rate of 15 mg/h. Blood samples were
taken at 8 and 24 hours, and plasma drug concentrations were
5.5 and 6.5 mg/L, respectively. Estimate the elimination half-
life of the drug in this patient?
If the desired therapeutic plasma concentration is 8 mg/L for
the above patient, what is the suitable infusion rate for the
patient?
From example, the trial infusion rate was 15 mg/h. Assuming
the second blood sample is the steady-state level, 6.5 mg/mL,
the clearance of the patient is
Css = R/Cl , Cl = R/Css = 15/6.5 = 2.31 L/h
The new infusion rate R= Css . Cl = 8 × 2.31 = 18.48 mg/h
15 mg/h = R2 R2 = 18.48 mg/h

6.5 mg/ mL 8 mg/mL
Summary:
 An IV drug infusion slowly inputs the drug into the
circulation and can provide stable drug concentrations (zero
order) in the plasma for extended time periods.
 Steady state is achieved when the rate of drug infusion

equals the rate of drug elimination
 Five elimination half-lives are needed to achieve 95% of

steady state.
plasma drug concentration at any time
during infusion
Steady-state drug concentration
t95% Css= 4.32 t1/2 Time need to reach 95%Css

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Intravenous

IV bolus injection IV infusion

 the duration of drug therapy may be maintained or

The change in the amount of drug in the body at any time

Cp is the plasma drug concentration at any time during the IV

At steady state, the rate of drug eliminaton from the body is

 No net change in the amount of drug in the body, DB, as a

Whenever the infusion stops, either before

For a zero-order elimination processes, if rate of input is

 In clinical practice, the drug activity will be observed

The time to reach steady-state

An increase in the infusion rate will not shorten the time

If the patient has a uremic condition and the elimination rate

Css = R/Cl , Cl = R/Css = 15/6.5 = 2.31 L/h

The new infusion rate R= Css . Cl = 8 × 2.31 = 18.48 mg/h

15 mg/h = R2 R2 = 18.48 mg/h

 Steady state is achieved when the rate of drug infusion

 Five elimination half-lives are needed to achieve 95% of

Steady-state drug concentration

t95% Css= 4.32 t1/2 Time need to reach 95%Css

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